Find $ \lim_{x\to 0}g(x)$ for $g(x)=\dfrac{4x^2-15}{4x+3}$.
Solution: $g$ is a rational function. Rational functions are continuous across their entire domain, and their domain is all real $x$ -values that don't make the denominator equal to zero. In other words, for any rational function $r$ and any input $c$ in the domain of $r$, we know that this equality holds: $\lim_{x\to c}r(x)=r(c)$ The input $x=0$ is within the domain of $g$. Therefore, in order to find $ \lim_{x\to 0}g(x)$, we can simply evaluate $g$ at $x=0$. $\begin{aligned} &\phantom{=}g(x) \\\\ &=\dfrac{4x^2-15}{4x+3} \\\\ &=\dfrac{4(0)^2-15}{4(0)+3} \gray{\text{Substitute }x=0} \\\\ &=\dfrac{-15}{3} \\\\ &=-5 \end{aligned}$ In conclusion, $ \lim_{x\to 0}g(x)=-5$.